y^2+y-(4x-16x^2)/10=0

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Solution for y^2+y-(4x-16x^2)/10=0 equation: 


x in (-oo:+oo)

t_1 = y^2+y

t_1-((4*x-16*x^2)/10) = 0

(-1*(4*x-16*x^2))/10+t_1 = 0

(-1*(4*x-16*x^2))/10+(10*t_1)/10 = 0

10*t_1-1*(4*x-16*x^2) = 0

10*t_1+16*x^2-4*x = 0

(10*t_1+16*x^2-4*x)/10 = 0

(10*t_1+16*x^2-4*x)/10 = 0 // * 10

10*t_1+16*x^2-4*x = 0

10*t_1+16*x^2-4*x = 0

DELTA = (-4)^2-(4*10*16*t_1)

DELTA = 16-640*t_1

16-640*t_1 = 0

16-640*t_1 = 0 // - 16

-640*t_1 = -16 // : -640

t_1 = -16/(-640)

t_1 = 1/40

DELTA = 0  <=>  t_4 = 1/40

x = 4/(2*16) i t_1 = 1/40

x = 1/8 i t_1 = 1/40

( x = ((16-640*t_1)^(1/2)+4)/(2*16) or x = (4-(16-640*t_1)^(1/2))/(2*16) ) i t_1 > 1/40

( x = ((16-640*t_1)^(1/2)+4)/32 or x = (4-(16-640*t_1)^(1/2))/32 ) i t_1 > 1/40

t_1-1/40 > 0

t_1-1/40 > 0 // + 1/40

t_1 > 1/40

x in { 1/8, ((16-640*(y^2+y))^(1/2)+4)/32, (4-(16-640*(y^2+y))^(1/2))/32 }

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